Use the rational zeros theorem to find all the real zeros of the polynomial function. Use the zeros to factor $f$ over the real numbers:
\[
f(x)=7 x^{3}-x^{2}+7 x-1
\]

Step 1 ：The Rational Root Theorem states that if a polynomial has a rational root, then it must be a factor of the constant term divided by a factor of the leading coefficient. In this case, the constant term is -1 and the leading coefficient is 7. The factors of -1 are -1 and 1, and the factors of 7 are -7, -1, 1, and 7. Therefore, the possible rational roots of the polynomial are \(\pm 1, \pm 1/7\).

Step 2 ：We can test these values by substituting them into the polynomial and checking if the result is zero. If the result is zero, then that value is a root of the polynomial.

Step 3 ：The roots of the polynomial are 1/7, -I, and I. However, the question asks for the real roots, so we can ignore the complex roots -I and I. Therefore, the only real root of the polynomial is 1/7. We can use this root to factor the polynomial.

Step 4 ：The polynomial \(f(x)\) can be factored as \((x - 1/7)(7x^2 + 7)\). Therefore, the real zeros of the polynomial are the solutions to the equations \(x - 1/7 = 0\) and \(7x^2 + 7 = 0\). We already know that \(x = 1/7\) is a solution to the first equation. The second equation has no real solutions because it is a quadratic equation with a negative discriminant. Therefore, the only real zero of the polynomial is \(x = 1/7\).

Step 5 ：Final Answer: The real zeros of the polynomial are \(\boxed{x = \frac{1}{7}}\). The polynomial can be factored over the real numbers as \(\boxed{f(x) = (x - \frac{1}{7})(7x^2 + 7)}\).